AMC 8 · 2013 · #10
Grade 6 number-theoryProblem
What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?
Pick an answer.
AMC 8 2013 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Find the ratio $\dfrac{\operatorname{lcm}(180, 594)}{\gcd(180, 594)}$.
Givens: Two whole numbers: $180$ and $594$; The ratio we want is LCM divided by GCF (also called GCD); Answer choices: (A) $110$, (B) $165$, (C) $330$, (D) $625$, (E) $660$
Unknowns: The whole-number ratio $\operatorname{lcm}(180, 594) \div \gcd(180, 594)$
Understand
Restated: Find the ratio $\dfrac{\operatorname{lcm}(180, 594)}{\gcd(180, 594)}$.
Givens: Two whole numbers: $180$ and $594$; The ratio we want is LCM divided by GCF (also called GCD); Answer choices: (A) $110$, (B) $165$, (C) $330$, (D) $625$, (E) $660$
Plan
Primary tool: #7 Identify Subproblems
Secondary: #9 Solve an Easier Related Problem
The question packs three jobs into one: find the GCF, find the LCM, then divide. Tool #7 (Identify Subproblems) makes those steps explicit — and we get all three from a single prime factorization of each number, so the work shares a foundation. Tool #9 (Easier Related Problem) supports the plan: trying a small case like $12$ and $18$ shows the clean shortcut $\dfrac{\operatorname{lcm}}{\gcd} = \dfrac{a \cdot b}{\gcd^2}$, which we can use as a check at the end. We prefer this concrete factor-tower approach over algebra (#13) because it keeps every step a small arithmetic move a Grade 6 student can verify.
Execute — Answer: C
4.OA.B.4 Step 1 - Subproblem 1: prime factorize $180$.
- Break it down repeatedly into smaller factors until only primes remain.
💡 Finding the prime building blocks of a whole number is the Grade 4 "factors and multiples" idea.
4.OA.B.4 Step 2 - Subproblem 2: prime factorize $594$.
- Same procedure — peel off small primes until none are left.
💡 Repeatedly dividing by the smallest prime that fits is the standard factoring loop.
6.NS.B.4 Step 3 - Subproblem 3: read off the GCF.
- For each prime that appears in BOTH lists, keep the smaller exponent.
- $2$: $\min(2,1) = 1$.
- $3$: $\min(2,3) = 2$.
- The primes $5$ and $11$ appear in only one list, so they are skipped.
💡 GCF = "what the two numbers share" — only common primes, and only as many copies as both have.
6.NS.B.4 Step 4 - Subproblem 4: read off the LCM.
- For each prime that appears in EITHER list, keep the larger exponent.
- $2$: $\max(2,1) = 2$.
- $3$: $\max(2,3) = 3$.
- $5$: only in $180$, take $5^1$.
- $11$: only in $594$, take $11^1$.
💡 LCM = "the smallest number both divide into" — every prime needs enough copies for both.
6.EE.A.1 Step 5 - Subproblem 5: form the ratio.
- Don't multiply the LCM out first — divide the prime towers directly, subtracting exponents on matching primes.
💡 Dividing same-base powers means subtracting exponents — easier than multiplying out $2^2 \cdot 3^3 \cdot 5 \cdot 11 = 5940$ and then dividing by $18$.
4.OA.B.4 Step 6 Multiply the four small primes to finish.
💡 A single multiplication chain finishes the problem with no big numbers in sight.
4.OA.B.4 Subproblem 1: prime factorize $180$. Break it down repeatedly into smaller facto 4.OA.B.4 Subproblem 2: prime factorize $594$. Same procedure — peel off small primes unti 6.NS.B.4 Subproblem 3: read off the GCF. For each prime that appears in BOTH lists, keep 6.NS.B.4 Subproblem 4: read off the LCM. For each prime that appears in EITHER list, keep 6.EE.A.1 Subproblem 5: form the ratio. Don't multiply the LCM out first — divide the prim 4.OA.B.4 Multiply the four small primes to finish. Review
Reasonableness: Use the identity $\operatorname{lcm}(a, b) \cdot \gcd(a, b) = a \cdot b$ as an independent check: $\dfrac{\operatorname{lcm}}{\gcd} = \dfrac{a \cdot b}{\gcd^2} = \dfrac{180 \cdot 594}{18^2} = \dfrac{180}{18} \cdot \dfrac{594}{18} = 10 \cdot 33 = 330$. Two different computations agree on $330$, so answer (C) is solid.
Alternative: Tool #9 (Solve an Easier Related Problem) — try the same question on $12$ and $18$ first. $12 = 2^2 \cdot 3$, $18 = 2 \cdot 3^2$, so $\gcd = 2 \cdot 3 = 6$ and $\operatorname{lcm} = 2^2 \cdot 3^2 = 36$. The ratio is $36 / 6 = 6 = 2 \cdot 3$, which is exactly the product of the primes that appear with *different* exponents. The same rule on $180$ and $594$ keeps the primes $2, 3, 5, 11$ and gives $330$. Same answer (C).
CCSS standards used (min grade 6)
4.OA.B.4Find all factor pairs for a whole number; recognize multiples and factors (Prime-factorizing $180 = 2^2 \cdot 3^2 \cdot 5$ and $594 = 2 \cdot 3^3 \cdot 11$, and finishing with the multiplication $2 \cdot 3 \cdot 5 \cdot 11 = 330$.)6.NS.B.4Find the greatest common factor and least common multiple of two whole numbers (Reading $\gcd(180, 594) = 18$ off the shared primes and $\operatorname{lcm}(180, 594) = 2^2 \cdot 3^3 \cdot 5 \cdot 11$ off all primes used by either number.)6.EE.A.1Write and evaluate numerical expressions involving whole-number exponents (Dividing the prime-power expressions by subtracting exponents on matching bases: $2^{2-1} \cdot 3^{3-2} \cdot 5 \cdot 11$.)
⭐ This AMC 8 problem just needs Grade 6 prime-factor reasoning — factor each number once, then read off the GCF and LCM from the same tower.
⭐ This AMC 8 problem just needs Grade 6 prime-factor reasoning — factor each number once, then read off the GCF and LCM from the same tower.
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